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Homework Help: Matrices Question

  1. Apr 8, 2007 #1
    I have a question about commutativity.
    I have two matrices X and Y and a constant k. I want to calculate X * kY. Can I bring k out the front to give k(X*Y)?
  2. jcsd
  3. Apr 9, 2007 #2
  4. Apr 9, 2007 #3
    Thanks for the link. It says "When the underlying ring is commutative, for example, the real or complex number field, the two multiplications are the same. However, if the ring is not commutative, such as the quaternions, they may be different."

    Lol, I am actually working with quaternions. The matrices contain complex numbers. I am trying to show that
    Q * Q-1 = Identity
    but Q-1 is kX because its the inverse of a 2x2 matrix. So I thought it would be easier to work out QX then multiply the answer by k to (hopefully) give the Identity matrix (if that makes any sense at all).
  5. Apr 9, 2007 #4
    Well then, I bow to people more knowledgeable than I about manipulating quarternions.
  6. Apr 9, 2007 #5


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    Same statement is true: a scalar (number), k, can be moved around pretty much as you wish. It is only multiplication of the matrices or quaternions that is non-commutative.
  7. Apr 9, 2007 #6
    Thanks for the help, I moved k out the front and it worked :smile:
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